Find $\lim_{x\to0} \frac{\log(1+3x)}{f(x)}$ given $\lim_{x\to0} \frac{f(x)}{\sin(x)} = 2$

How about using these (well-known) limits:

$$\lim_{x\to 0} \frac{\ln (1+x)}{x}=1,\ \ \lim_{x\to 0} \frac{\sin x}{x}=1$$

and writing:

$$\lim_{x\to 0}\frac{\ln(1+3x)}{f(x)} = 3\lim _{x\to 0} \left[\frac{\ln (1+3x)}{3x}\cdot \frac{x}{\sin x}\cdot \frac{\sin x}{f(x)}\right]$$

Can you end it from here?

Hint:

If the limits exist,

$$\lim_{x\to0} \frac{f(x)}{\sin(x)}\cdot\lim_{x\to0} \frac{\log(1+3x)}{f(x)}=\lim_{x\to0} \frac{f(x)}{\sin(x)}\frac{\log(1+3x)}{f(x)}=\lim_{x\to0} \frac{\log(1+3x)}{\sin(x)}.$$

With equivalence of functions near $0$:

The hypothesis means that $f(x)\sim_0 2\sin x$

On the other hand $\sin x\sim_0 x$ and $\ln(1+x)\sim_0 x$, so $$\frac{\log(1+3x)}{f(x)}\sim_0 \frac{3x}{2x}=\frac 32.$$